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Saudi Arabian IMO Booklet

Geometry Difficulty 8.6 Shortlist Prove it Saudi Arabia

Given is triangle $ABC$ with $AB > AC$ . Circles $o_B$ , $o_C$ are inscribed in angle $BAC$ with $o_B$ tangent to $AB$ at $B$ and $o_C$ tangent to $AC$ at $C$ . Tangent to $o_B$ from $C$ different than $AC$ intersects $AB$ at $K$ , and tangent to $o_C$ from $B$ different than $AB$ intersects $AC$ at $L$ . Line $KL$ and the angle bisector of $BAC$ intersect $BC$ at points $P$ and $M$ , respectively. Prove that $BP = CM$ .

Solution

Note that the length of the segment of the common tangent to $o_B$ and $o_C$ joining the tangency points is equal to
$AB - AC = LB - LC = KC - KB,$
SAUDI ARABIAN IMO Booklet 2022

which means that points $A$ , $L$ lie on the one, and point $K$ on the other leg of some hyperbola $\eta$ with foci $B$ , $C$ .
Denote by $S$ the midpoint of the segment $BC$ (the center of symmetry of $\eta$ ) and denote by $A'$ , $K'$ , $M'$ the central reflections in $S$ of $A$ , $K$ , $M$ , respectively. Then $A'$ , $K' \in \eta$ and $A'$ , $K'$ , $C$ are collinear.

From the optical property of a hyperbola follows that $A'M'$ is tangent to $\eta$ (as it is the bisector of $BA'C$ ). Therefore Pascal's theorem applied to the degenerate hexagon $AA'A'K'KL$ inscribed in $\eta$ gives the collinearity of points
$AA' \cap KK' = S, \quad A'A' \cap KL, \quad A'K' \cap LA = C.$
Therefore the three lines $A'A'$ (tangent to $\eta$ in $A'$ ), $KL$ , $BC$ are concurrent, which means that $M' = A'A' \cap BC = KL \cap BC = P$ and in consequence $CM = BP$ . $\square$

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